What is the hybridization of the iodine atom in ICl_3?

• A. sp^3
• B. sp^2d
• C. sp^2
• D. sp^3d

Answer: (D)

Verified step by step guidance

1. Step 1: Determine the total number of valence electrons around the iodine atom. Iodine (I) has 7 valence electrons, and each chlorine (Cl) atom also contributes 7 valence electrons, but since we focus on the central atom's hybridization, start with iodine's 7 electrons and consider bonding with 3 chlorine atoms.
2. Step 2: Draw the Lewis structure of ICl_3. Place iodine in the center bonded to three chlorine atoms. Each I–Cl bond uses 2 electrons, so 3 bonds use 6 electrons. Subtract these from iodine's 7 valence electrons to find the remaining electrons as lone pairs on iodine.
3. Step 3: Count the regions of electron density (electron domains) around iodine. Each bond counts as one region, and each lone pair counts as one region. The total number of electron domains determines the electron geometry and hybridization.
4. Step 4: Use the number of electron domains to assign the hybridization. For example, 5 electron domains correspond to sp^3d hybridization because one s, three p, and one d orbital mix to accommodate five regions of electron density.
5. Step 5: Confirm that the hybridization of iodine in ICl_3 is sp^3d, which corresponds to a trigonal bipyramidal electron geometry with lone pairs occupying equatorial positions, consistent with the molecular shape of ICl_3.